Question

The equation of the parabola whose focus is $(-6,-6)$ and vertex is $(-2,2),$ is

• A. $(2x-y{)}^2+104x+148y-124=0$
• B. $(2x-y{)}^2+104x-148y+124=0$
• C. $(2x-y{)}^2-104x+148y+124=0$
• D. $(2x+y{)}^2+104x+148y-124=0$

Answer 1

The correct option is A $(2x-y{)}^2+104x+148y-124=0$


$\because A$ is mid point of $SK,$
$\Rightarrow A(-2,2)=(\frac{x_1-6}{2},\frac{y_1-6}{2})$; where $S=(-6,-6)$ and $K=(x_1,y_1)$
$\Rightarrow -2=\frac{x_1-6}{2},2=\frac{y_1-6}{2}$
$\Rightarrow x_1-6=-4$, $y_1-6=4$
$\therefore x_1=2,y_1=10$
Hence, coordinates of $K\equiv (2,10)$
Slope of SK$=\frac{-6-2}{-6+2}=\frac{-8}{-4}=2$
Now, directrix of the required parabola will be $\perp SK$
So, slope SK is $=-\frac{1}{2}$
Hence, equation of the directrix will be
$y-10=(-\frac{1}{2})(x-2)$
$\Rightarrow 2y-20=-x+2$
$\Rightarrow x+2y=22$
The distance between S and P is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Where $(x_1,y_1)=(-6,-6)$ and $(x_2,y_2)=(x,y)$
So, $SP=\sqrt{(x+6{)}^2+(y+6{)}^2}$
Similar way, the distance between point $P(x,y)$ and the line $x+2y=22$, we get get $PM$ distance
So, $PM=\frac{|x+2y-22|}{\sqrt{1^2+2^2}}$.
Now, for parabola condition $SP=PM$
$\sqrt{(x+6{)}^2+(y+6{)}^2}=\frac{|x+2y-22|}{\sqrt{1^2+2^2}}$
Squaring on both sides, we get
$(x+6{)}^2+(y+6{)}^2=\frac{(x+2y-22{)}^2}{1+4}$
$\Rightarrow 5(x^2+y^2+12x+12y+36+36)=(x+2y-22{)}^2$
$\Rightarrow 5x^2+5y^2+60x+60y+360=x^2+4y^2+484+4xy-88y-44x$
$\Rightarrow 5x^2+5y^2+60x+60y+360-x^2-4y^2-4xy+88y+44x-484=0$
$\Rightarrow 4x^2+y^2-4xy+104x+148y-124=0\Rightarrow (2x-y{)}^2+104x+148y-124=0$